step 1: find molar mass use: Tf = Kf*mb 0.73 = 1.99 *mb mb= 0.3668 molal m(solve
ID: 699456 • Letter: S
Question
step 1: find molar mass
use:
Tf = Kf*mb
0.73 = 1.99 *mb
mb= 0.3668 molal
m(solvent)= 100 g
= 0.1 kg
use:
number of mol,
n = Molality * mass of solvent in Kg
= (0.3668 mol/Kg)*(0.1 Kg)
= 3.668*10^-2 mol
mass(solute)= 5.18 g
use:
number of mol = mass / molar mass
3.668*10^-2 mol = (5.18 g)/molar mass
molar mass = 141.2 g/mol
step 2: find empirical formula and then molecular formula
let in compound number of moles of C, H and O be x, y and z respectively
Number of moles of CO2 = mass of CO2 / molar mass CO2
= 4.846/44
= 0.1101
Number of moles of H2O = mass of H2O / molar mass H2O
= 1.488/18
= 0.0827
Since 1 mol of CO2 has 1 mol of C
Number of moles of C in CO2= 0.1101
so, x = 0.1101
Since 1 mol of H2O has 2 mol of H
Number of moles of H = 2*0.0827 = 0.1653
Molar mass of O = 16 g/mol
mass O = total mass - mass of C and H
= 2.37 - 0.1101*12 - 0.1653*1
= 0.883
number of mol of O = mass of O / molar mass of O
= 0.883/16.0
= 0.0552
so, z = 0.0552
Divide by smallest to get simplest whole number ratio:
C: 0.1101/0.0552 = 2
H: 0.1653/0.0552 = 3
O: 0.0552/0.0552 = 1
So empirical formula is:C2H3O
Molar mass of C2H3O,
MM = 2*MM(C) + 3*MM(H) + 1*MM(O)
= 2*12.01 + 3*1.008 + 1*16.0
= 43.044 g/mol
Now we have:
Molar mass = 141.2 g/mol
Empirical formula mass = 43.044 g/mol
Multiplying factor = molar mass / empirical formula mass
= 141.2/43.044
= 3
So molecular formula is:C6H9O3
Explanation / Answer
An unknown compound contains carbon, hydrogen and oxygen. Comibustion analysis of a 2.37g mple produced 4.846g of CO2 and 1.488g of H:O. In a separate experiment, a S sa was dissolved in 100g of ethanol, CH,CH OH. The solution froze at -115.4"C. The freezing point of pure ethamol is-114.67 and it has a Kt of 195C kg mol. what is the molecular formula of the compound? Show your work for full credit.
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